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  • Computer Science graduate. Master or full-time job? [closed]

    - by Alex
    Possible Duplicate: Is a Master's worth it? I have just gotten my Bachelor's Degree in Computer Science and I have to make choice. Whether to continue with my full-time job I just got or put the job slightly in the background and concentrate on getting a Master's degree. I am currently working as an embedded C developer in a small company. The cool thing is that, because the team is quite small, my engineering ideas really play a part in the final product. Not to mention that I get to work on very different areas of embedded programming: device drivers and development of a Real Time OS. I am very enthusiastic about my job and what I do. On the other hand, in my country there isn't really a master's degree that focuses on embedded development so my gain from getting this degree will mainly in the field of general computer science knowledge. That being said, is it worth giving up all my spare time which I now use to study different areas of embedded devices and work mainly to get a degree rather than pure knowledge and experience in the field I want to work in?

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  • make a thread which recieves values from other threads

    - by farteaga88
    This program in Java creates a list of 15 numbers and creates 3 threads to search for the maximum in a given interval. I want to create another thread that takes those 3 numbers and get the maximum. but i don't know how to get those values in the other thread. public class apple implements Runnable{ String name; int time, number, first, last, maximum; int[] array = {12, 32, 54 ,64, 656, 756, 765 ,43, 34, 54,5 ,45 ,6 , 5, 65}; public apple(String s, int f, int l){ name = s; first = f; last = l; maximum = array[0]; } public void run(){ try{ for(int i = first; i < last; i++ ) { if(maximum < array[i]) { maximum = array[i]; } } System.out.println("Thread"+ name + "maximum = " + maximum); }catch(Exception e){} } public static void main(String[] args){ Thread t1 = new Thread(new apple("1 ", 0, 5)); Thread t2 = new Thread(new apple("2 ", 5, 10 )); Thread t3 = new Thread(new apple("3 ", 10, 15)); try{ t1.start(); t2.start(); t3.start(); }catch(Exception e){} } }

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  • Semaphore - What is the use of initial count?

    - by Sandbox
    http://msdn.microsoft.com/en-us/library/system.threading.semaphoreslim.aspx To create a semaphore, I need to provide an initial count and maximum count. MSDN states that an initial count is - The initial number of requests for the semaphore that can be granted concurrently. While it states that maximum count is The maximum number of requests for the semaphore that can be granted concurrently. I can understand that the maximum count is the maximum number of threads that can access a resource concurrently. But, what is the use of initial count? If I create a semaphore with an initial count of 0 and a maximum count of 2, none of my threadpool threads are able to access the resource. If I set the initial count as 1 and maximum count as 2 then only thread pool thread can access the resource. It is only when I set both initial count and maximum count as 2, 2 threads are able to access the resource concurrently. So, I am really confused about the significance of initial count? SemaphoreSlim semaphoreSlim = new SemaphoreSlim(0, 2); //all threadpool threads wait SemaphoreSlim semaphoreSlim = new SemaphoreSlim(1, 2);//only one thread has access to the resource at a time SemaphoreSlim semaphoreSlim = new SemaphoreSlim(2, 2);//two threadpool threads can access the resource concurrently

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  • Efficient algorithm to find a maximum common subset of two sets?

    - by datasunny
    Each set contains bunch of checksums. For example: Set A: { 4445968d0e100ad08323df8c895cea15 a67f8052594d6ba3f75502c0b91b868f 07736dde2f8484a4a3af463e05f039e3 5b1e374ff2ba949ab49870ca24d3163a } Set B: { 6639e1da308fd7b04b7635a17450df7c 4445968d0e100ad08323df8c895cea15 a67f8052594d6ba3f75502c0b91b868f } The maximum common subset of A and B is: { 4445968d0e100ad08323df8c895cea15 a67f8052594d6ba3f75502c0b91b868f } A lot of this operations will be performed, so I'm looking for an efficient algorithm to do so. Thanks for your help.

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  • Finding the maximum weight subsequence of an array of positive integers?

    - by BeeBand
    I'm tring to find the maximum weight subsequence of an array of positive integers - the catch is that no adjacent members are allowed in the final subsequence. The exact same question was asked here, and a recursive solution was given by MarkusQ. He provides an explanation, but can anyone help me understand how he has expanded the function? How does this solution take into consideration non-adjacent members?

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  • What's the maximum number of operators you that can be used in an if statement?

    - by DMinGod
    Hi, i'm using jquery and I'm trying to validate a form. My question is - What is the maximum number of tests can you give in a single if statement. function cc_validate () { if ($("#name").val() == "" || $("#ship_name").val() == "" || $("#address").val() == "" || $("#city").val() == "" || $("#ship_city").val() == "" || $("#state").val() == "" || $("#ship_state").val() == "" || $("#postal_code").val() == "" || isNaN($("#postal_code").val()) || $("#phone").val() == "" || $("#ship_phone").val() == "" || isNaN($("#phone").val()) || isNaN($("#ship_phone").val()) || $("#mobile_number").val() == "" || $("#ship_mobile_number").val() == "" || isNaN($("#mobile_number").val()) || isNaN($("#ship_mobile_number").val()) || $("#email").val() == "") { return false; } else { return true; } }

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  • How to select product that have the maximum price of each category?

    - by kimleng
    The below is my table that has the item such as: ProductId ProductName Category Price 1 Tiger Beer $12.00 2 ABC Beer $13.99 3 Anchor Beer $9.00 4 Apolo Wine $10.88 5 Randonal Wine $18.90 6 Wisky Wine $30.19 7 Coca Beverage $2.00 8 Sting Beverage $5.00 9 Spy Beverage $4.00 10 Angkor Beer $12.88 And I suppose that I have only three category in this table (I can have a lot of category in this table). And I want to show the maximum product's price of each category in this table.

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  • What's the actual windows media player maximum banner width?

    - by William
    When looking at the documentation for the banner tag in windows media metafiles all the documentation says there is a maximum size of 194 x 32 pixels. http://msdn.microsoft.com/en-us/library/dd562478(v=VS.85).aspx However, when experimenting with larger images they seem to be displaying correctly. Is there some real undocumented limitation here? Will it get cut off on different/older versions?

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  • Min-Ordered Bionomial Heap Insertion java

    - by Charodd Richardson
    Im writing a java code to make a min-ordered Binomial Heap and I have to Insert and Remove-min. I'm having a very big problem inserting into the Heap. I have been stuck on this for a couple of days now and it is due tomorrow. Whenever I go to insert, It only prints out the item I insert instead of the whole tree (which is in preorder). Such as if I insert 1 it prints (1) and then I go to insert 2 it prints out (2) instead of (1(2)) It keeps printing out only the number I insert last instead of the whole preordered tree. I would be very grateful if someone could help me with this problem. Thank you so much in advance, Here is my code. public class BHeap { int key; int degree;//The degree(Number of children) BHeap parent, leftmostChild, rightmostChild, rightSibling,root,previous,next; public BHeap(){ key =0; degree=0; parent =null; leftmostChild=null; rightmostChild=null; rightSibling=null; root=null; previous=null; next=null; } public BHeap merge(BHeap x, BHeap y){ BHeap newHeap = new BHeap(); y.rightSibling=x.root; BHeap currentHeap = y; BHeap nextHeap = y.rightSibling; while(currentHeap.rightSibling !=null){ if(currentHeap.degree==nextHeap.degree){ if(currentHeap.key<nextHeap.key){ if(currentHeap.degree ==0){ currentHeap.leftmostChild=nextHeap; currentHeap.rightmostChild=nextHeap; currentHeap.rightSibling=nextHeap.rightSibling; nextHeap.rightSibling=null; nextHeap.parent=currentHeap; currentHeap.degree++; } else{ newHeap = currentHeap; newHeap.rightmostChild.rightSibling=nextHeap; newHeap.rightmostChild=nextHeap; nextHeap.parent=newHeap; newHeap.degree++; nextHeap.rightSibling=null; nextHeap=newHeap.rightSibling; } } else{ if(currentHeap.degree==0){ nextHeap.rightmostChild=currentHeap; nextHeap.rightmostChild.root = nextHeap.rightmostChild;//add nextHeap.leftmostChild=currentHeap; nextHeap.leftmostChild.root = nextHeap.leftmostChild;//add currentHeap.parent=nextHeap; currentHeap.rightSibling=null; currentHeap.root=currentHeap;//add nextHeap.degree++; } else{ newHeap=nextHeap; newHeap.rightmostChild.rightSibling=currentHeap; newHeap.rightmostChild=currentHeap; currentHeap.parent= newHeap; newHeap.degree++; currentHeap=newHeap.rightSibling; currentHeap.rightSibling=null; } } } else{ currentHeap=currentHeap.rightSibling; nextHeap=nextHeap.rightSibling; } } return y; } public void Insert(int x){ /*BHeap newHeap = new BHeap(); newHeap.key=x; if(this.root==null){ this.root=newHeap; return; } else{ this.root=merge(newHeap,this.root); }*/ BHeap newHeap= new BHeap(); newHeap.key=x; if(this.root==null){ this.root=newHeap; } else{ this.root = merge(this,newHeap); }} public void RemoveMin(){ BHeap newHeap = new BHeap(); BHeap child = new BHeap(); newHeap=this; BHeap pos = newHeap.next; while(pos !=null){ if(pos.key<newHeap.key){ newHeap=pos; } pos=pos.rightSibling; } pos=this; BHeap B1 = new BHeap(); if(newHeap.previous!=null){ newHeap.previous.rightSibling=newHeap.rightSibling; B1 =pos.leftmostChild; B1.rightSibling=pos; pos.leftmostChild=pos.rightmostChild.leftmostChild; } else{ newHeap=newHeap.rightSibling; newHeap.previous.rightSibling=newHeap.rightSibling; B1 =pos.leftmostChild; B1.rightSibling=pos; pos.leftmostChild=pos.rightmostChild.leftmostChild; } merge(newHeap,B1); } public void Display(){ System.out.print("("); System.out.print(this.root.key); if(this.leftmostChild != null){ this.leftmostChild.Display(); } System.out.print(")"); if(this.rightSibling!=null){ this.rightSibling.Display(); } } }

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  • How to find minimum weight with maximum cost in 0-1 Knapsack algorithm?

    - by Nitin9791
    I am trying to solve a spoj problem Party Schedule the problem statement is- You just received another bill which you cannot pay because you lack the money. Unfortunately, this is not the first time to happen, and now you decide to investigate the cause of your constant monetary shortness. The reason is quite obvious: the lion's share of your money routinely disappears at the entrance of party localities. You make up your mind to solve the problem where it arises, namely at the parties themselves. You introduce a limit for your party budget and try to have the most possible fun with regard to this limit. You inquire beforehand about the entrance fee to each party and estimate how much fun you might have there. The list is readily compiled, but how do you actually pick the parties that give you the most fun and do not exceed your budget? Write a program which finds this optimal set of parties that offer the most fun. Keep in mind that your budget need not necessarily be reached exactly. Achieve the highest possible fun level, and do not spend more money than is absolutely necessary. Input The first line of the input specifies your party budget and the number n of parties. The following n lines contain two numbers each. The first number indicates the entrance fee of each party. Parties cost between 5 and 25 francs. The second number indicates the amount of fun of each party, given as an integer number ranging from 0 to 10. The budget will not exceed 500 and there will be at most 100 parties. All numbers are separated by a single space. There are many test cases. Input ends with 0 0. Output For each test case your program must output the sum of the entrance fees and the sum of all fun values of an optimal solution. Both numbers must be separated by a single space. Example Sample input: 50 10 12 3 15 8 16 9 16 6 10 2 21 9 18 4 12 4 17 8 18 9 50 10 13 8 19 10 16 8 12 9 10 2 12 8 13 5 15 5 11 7 16 2 0 0 Sample output: 49 26 48 32 now I know that it is an advance version of 0/1 knapsack problem where along with maximum cost we also have to find minimum weight that is less than a a given weight and have maximum cost. so I have used dp to solve this problem but still get a wrong awnser on submission while it is perfectly fine with given test cases. My code is typedef vector<int> vi; #define pb push_back #define FOR(i,n) for(int i=0;i<n;i++) int main() { //freopen("input.txt","r",stdin); while(1) { int W,n; cin>>W>>n; if(W==0 && n==0) break; int K[n+1][W+1]; vi val,wt; FOR(i,n) { int x,y; cin>>x>>y; wt.pb(x); val.pb(y); } FOR(i,n+1) { FOR(w,W+1) { if(i==0 || w==0) { K[i][w]=0; } else if (wt[i-1] <= w) { if(val[i-1] + K[i-1][w-wt[i-1]]>=K[i-1][w]) { K[i][w]=val[i-1] + K[i-1][w-wt[i-1]]; } else { K[i][w]=K[i-1][w]; } } else { K[i][w] = K[i-1][w]; } } } int a1=K[n][W],a2; for(int j=0;j<W;j++) { if(K[n][j]==a1) { a2=j; break; } } cout<<a2<<" "<<a1<<"\n"; } return 0; } Could anyone suggest what am I missing??

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  • Android: How to set the maximum size of a Spinner?

    - by Epaga
    Here is my layout <?xml version="1.0" encoding="utf-8"?> <LinearLayout xmlns:android="http://schemas.android.com/apk/res/android" android:layout_width="fill_parent" android:id="@+id/LinearLayout01" android:layout_height="wrap_content"> <Spinner android:text="@+id/AutoCompleteTextView01" android:id="@+id/Spinner01" android:layout_width="wrap_content" android:layout_height="wrap_content" android:width="130dp"></Spinner> <EditText android:layout_height="wrap_content" android:layout_width="wrap_content" android:id="@+id/Chapter" android:width="30dp"></EditText> <TextView android:layout_width="wrap_content" android:layout_height="wrap_content" android:id="@+id/TextView01" android:text=":"></TextView> <EditText android:layout_height="wrap_content" android:layout_width="wrap_content" android:id="@+id/Verse" android:width="40dp"></EditText> </LinearLayout> I inflate this layout as an AlertDialog's view. But when I pick a large element, the text fields get pushed out to the right. Is there any way to set the maximum size of the spinner so after choosing an element, it shortens the choice with an ellipsis ("...") or something?

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